(c+2)(2c-1)=

Solution for (c+2)(2c-1)= equation:

Simplifying
(c + 2)(2c + -1) = 0

Reorder the terms:
(2 + c)(2c + -1) = 0

Reorder the terms:
(2 + c)(-1 + 2c) = 0

Multiply (2 + c) * (-1 + 2c)
(2(-1 + 2c) + c(-1 + 2c)) = 0
((-1 * 2 + 2c * 2) + c(-1 + 2c)) = 0
((-2 + 4c) + c(-1 + 2c)) = 0
(-2 + 4c + (-1 * c + 2c * c)) = 0
(-2 + 4c + (-1c + 2c2)) = 0

Combine like terms: 4c + -1c = 3c
(-2 + 3c + 2c2) = 0

Solving
-2 + 3c + 2c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(-2 + -1c)(1 + -2c) = 0

Subproblem 1
Set the factor '(-2 + -1c)' equal to zero and attempt to solve:

Simplifying
-2 + -1c = 0

Solving
-2 + -1c = 0

Move all terms containing c to the left, all other terms to the right.

Add '2' to each side of the equation.
-2 + 2 + -1c = 0 + 2

Combine like terms: -2 + 2 = 0
0 + -1c = 0 + 2
-1c = 0 + 2

Combine like terms: 0 + 2 = 2
-1c = 2

Divide each side by '-1'.
c = -2

Simplifying
c = -2
Subproblem 2
Set the factor '(1 + -2c)' equal to zero and attempt to solve:

Simplifying
1 + -2c = 0

Solving
1 + -2c = 0

Move all terms containing c to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + -2c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -2c = 0 + -1
-2c = 0 + -1

Combine like terms: 0 + -1 = -1
-2c = -1

Divide each side by '-2'.
c = 0.5

Simplifying
c = 0.5
Solution
c = {-2, 0.5}